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Physics Question on Alternating current

An ideal capacitor of capacitance 0.2 μF0.2 \, \mu F is charged to a potential difference of 10 V.10\, V. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance 0.5 mH.0.5\, mH. The current at a time when the potential difference across the capacitor is 5 V,5\, V, is :

A

0.34 A0.34\, A

B

0.25 A0.25\, A

C

0.17 A0.17\, A

D

0.15 A0.15\, A

Answer

0.17 A0.17\, A

Explanation

Solution

The given circuit is

Given: capacitor is charged to a potential difference of 10 V⇒V0=10 V10\, V \Rightarrow V_{0}=10\, V
Therefore, charge on capacitor, Q0=CV0=0.2 μF×10 VQ_{0}=C V_{0}=0.2\, \mu F \times 10\, V
Now, Ec=E_{c}= Energy stored in capacitor =12CV02=\frac{1}{2} C V_{0}^{2}
⇒EC=12×0.2μF×(10V)2=10μJ=10×10−6 J\Rightarrow E_{C}=\frac{1}{2} \times 0.2 \mu F \times(10 V )^{2}=10 \mu J =10 \times 10^{-6}\, J
When inductor is connected in parallel with the capacitor in the circuit, the energy stored in capacitor when potential difference across capacitor is V′=5KV'=5 K is
EC′=12CV2=12×0.2μF×(5V)2E_{C}'=\frac{1}{2} C V^{2}=\frac{1}{2} \times 0.2 \mu F \times(5 V )^{2}
=2.5μ J=2.5×10−6 J=2.5 \mu\, J =2.5 \times 10^{-6}\, J
Therefore, Energy stored in inductor =Ec−Ec′=10×10−6J−2.5×10−6 J=E_{c}-E_{c}'=10 \times 10^{-6} J -2.5 \times 10^{-6}\, J
=7.5×10−6 J=7.5 \times 10^{-6}\, J
Also, we know energy stored in inductor =12LI2=\frac{1}{2} L I^{2}
⇒12LI2=7.5×10−6\Rightarrow \frac{1}{2} L I^{2}=7.5 \times 10^{-6}
⇒I2=7.5×10−6×20.5 mH=30×10−3\Rightarrow I^{2}=\frac{7.5 \times 10^{-6} \times 2}{0.5\, mH }=30 \times 10^{-3}
⇒I=310=0.17 A\Rightarrow I=\frac{\sqrt{3}}{10}=0.17\, A